[Haskell-cafe] bit of a noob question

[Jeremy Shaw]

There are many ways you can do it. Here are two. The first uses the  
Transform List Comp extensions introduced in 6.10.

http://www.haskell.org/ghc/docs/latest/html/users_guide/syntax-extns.html#generalised-list-comprehensions

The second uses more normal Haskell. The second version is probably  
not the best 'normal' Haskell implementation though.

{-# LANGUAGE TransformListComp #-}

import Data.Function (on)
import Data.List (groupBy)
import GHC.Exts

test :: (Ord a) => [(a, b)] -> [(a, [b])]
test l = [ (the f, s) | (f,s) <- l , then group by f ]

ex1 = test [('a',1),('a',2),('a',3),('b',1),('b',2)]


test2 :: (Ord a) => [(a, b)] -> [(a, [b])]
test2 l =
     map (\grp -> (fst (head grp), map snd grp)) ((groupBy ((==) `on`  
fst)) l)

ex2 = test [('a',1),('a',2),('a',3),('b',1),('b',2)]

On Oct 24, 2009, at 5:27 PM, spot135 wrote:

>
>
>
> Ok maybe a noob question, but hopefully its an easy one.
>
> This is what I've got so far:
>
> test :: x->[a] -> (b,[b])
> test x arrlist = let test1 = x
> 		         a = filter (\n -> fst n == test1) arrlist
> 		        test2 = map snd a
> 	           in (test1, [test2])
>
> so basically I have a list say [(a,1),(a,2),(a,3),(b,1),(b,2)] etc
> So I give the function a x value (a or b) in this case and it return
> (a,[1,2,3])
>
> which is all gravy
>
> But,
> Is there a way that i dont have to supply the a or b ie i call the  
> function
> and it gives me the list
> [(a,[1,2,3]),(b,[1,2])...
>
> I presume i need another layer of recursion but I cant figure out  
> how to do
> it.
>
> Any help would be gratefully received :-)
>
> -- 
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